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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    In the Bohr's model of hydrogen-like atom the force between the nucleus and the electron is modified as \[F=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right),\] where \[\beta \] is a constant. For this atom, the radius of the nth orbit in terms of the Bohr radius \[\left( {{a}_{0}}=\frac{{{\varepsilon }_{0}}{{h}^{2}}}{m\pi {{e}^{2}}} \right)\] is:

    A)  \[{{r}_{n}}={{a}_{0}}n-\beta \]

    B)  \[{{r}_{n}}={{a}_{0}}{{n}^{2}}+\beta \]

    C)  \[{{r}_{n}}={{a}_{0}}{{n}^{2}}-\beta \]

    D)  \[{{r}_{n}}={{a}_{0}}n+\beta \]

    Correct Answer: C

    Solution :

    [c] As \[F=\frac{m{{v}^{2}}}{r}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] \[\text{and }mvr=\frac{nh}{2\pi }\Rightarrow v=\frac{nh}{2\pi mr}\] \[\therefore m{{\left( \frac{nh}{2\pi mr} \right)}^{2}}\times \frac{1}{r}=\frac{{{e}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}} \right)\] or, \[\frac{{{a}_{0}}{{n}^{2}}}{{{r}^{3}}}=\frac{1}{{{r}^{2}}}+\frac{\beta }{{{r}^{3}}}\left( \therefore {{a}_{0}}=\frac{{{\varepsilon }_{0}}{{h}^{2}}}{m\pi {{e}^{2}}}\text{Given} \right)\] \[\therefore r={{a}_{0}}{{n}^{2}}-\beta \]       


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